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Home  >>  JEEMAIN and NEET  >>  Physics  >>  Class11  >>  Motion in a Straight Line

A train starts from station A with uniform acceleration $\alpha$ for some distance and then goes with uniform retardation $\beta$ for some more time to come to rest at station B.The distance between station A and B is 4 km. and the train takes 4 minutes to complete this journey. If $\alpha$ and $\beta$ are in $km/min^2$ then

\[(a)\;\frac{1}{\alpha}+\frac{1}{\beta}=2\quad (b)\;\frac{1}{\alpha}+\frac{1}{\beta}=4\quad (c)\;\frac{1}{\alpha}+\frac{1}{\beta}=\frac{1}{2} \quad (d)\;\frac{1}{\alpha}+\frac{1}{\beta}=\frac{1}{4}\]

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1 Answer

Since $S_1+S_2=4 km$
and $t_1+t_2=4 min$
$t_1=\large\frac{v_{\Large max}}{\alpha}$
$t_2=\large\frac{v_{\Large max}}{\beta}$
$t_1+t_2=v_{max}\bigg(\large\frac{1}{\alpha}+\frac{1}{\beta}\bigg)$
$S_1=\large\frac{1}{2} v_{max} t_1$
$S_2=\large\frac{1}{2} v_{max} t_2$
$=>v_{\large max}=2$
Therefore $\large\frac{1}{\alpha}+\frac{1}{\beta}$$=2$
Hence a is the correct answer. 

 

answered Jun 28, 2013 by meena.p
edited Jan 25, 2014 by meena.p
 

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